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I'm trying to show that for a first countable space $X$, that if $x$ is in the closure of a subset $A\subseteq X$, then $x$ is the limit of a sequence in $A$. In the process of doing so, I keep finding myself needing a countable local basis $\{B_n\}_{n=1}^\infty$ at $x$ which is linearly ordered by inclusion, i.e. such that $$B_1\supseteq B_2 \supseteq \cdots.$$ For the life of me, I can't seem to figure out how to find such a basis. How do I go about finding one/showing one exists?

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up vote 3 down vote accepted

Given any countable local base $\{B_n:n\in\Bbb Z^+\}$ at $x$, let $V_n=\bigcap_{k=1}^nB_k$, and use the local base $\{V_n:n\in\Bbb Z^+\}$ instead.

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I had a feeling it would end up being something simple. -sigh- – Avi Steiner Jul 12 '12 at 1:38

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